DISINI solutions2000

S

OLUTIONS

TO

USC’

S

2000 H

IGH

S

CHOOL

M

ATH

C

ONTEST

1. (a) Since2 3

3 2

x 2

=72x 2

=9800 =2 3

5 2

7 2

, we obtainx 2

=5 2

7 2

=3 2

. Sincex>0, we deduce that x=57=3=35=3.

2. (b) Sincelog 2

(log 2

(log 2

16))=log 2

(log 2

4)=log 2

2=1, the answer is1.

3. (e) Using that2 600

= 64 100

,3 500

=243 100

,4 400

=256 100

,5 300

=125 100

, and6 200

= 36 100

, we see that the answer is6

200

.

4. (b) Letx=35811231and observe that0<x<10 8

. Hence,

10

14 +x 10

14

2 =

10 28

+(2x)10 14

+x 2 10

28

=

10000007162246210 14

+x 2 10

28

:

Since0<x 2

<10 16

, we see that the expressionx 2

does not effect the leading 12 digits of the integer in the numerator. Thus,1:00000035811231

2

=1:000000716224:::. Hence,x=7,y=1, andz=6so thatx+y+z=14.

5. (e) Given

(x 3

x 2

5x 2)(x 4

+x 3

+kx 2

5x+2)=x 7

4x 5

14x 4

5x 3

+19x 2

4;

the coefficient ofx 2

on the left is 2k+25 2= 2k+23and the coefficient ofx 2

on the right is19. These must

be equal so that 2k+23=19. Hence,k=2.

6. (d) The 100 horizontal lines divide the plane into 101 regions. If each vertical line is considered in turn, each divides the plane into 101 additional regions. The total number of regions obtained is101

2

=10201.

7. (c) The answer is2 15

=32768since it is 2 3

5

=8 5

as well as 2 5

3

=32 3

. To see thatN cannot be smaller,

observe that the conditions in the problem implyN = a 5

for some positive integeraand that for1 a 7 the

numbera 5

is not a cube of a number different froma.

8. (d) Recall that2sinos=sin (2). Hence,

(sin15 Æ

) 2

(os15 Æ

) 2

=(sin15 Æ

os15 Æ

) 2

=

sin30 Æ 2

2

=

1 4

2

= 1 16

:

9. (a) Letx=BC. SinceDBCis a45-90-45degree triangle,DB=x. SinceABCis a30-90-60degree triangle

andtan30 Æ

=1= p

3, we deduce that 1 p 3

=tan30 Æ

= BC AB

= BC AD+DB

= x 2+x

:

Therefore,2+x=x p

3, from which we deduce thatBC=x=2=( p

3 1)= p

3+1.

10. (b) Since9 3

=729and10 3

=1000, we see that 3 p

829=9+rwhere0<r<1. Since10 2

=100and10 3

=1000,

we see thatlog 10

829=2+swhere0<s<1. Thus, 3

p

829+log 10

829 2

=

(9+r)+(2+s) 2

=

11+(r+s) 2

=5:5+t;

where0 <t =(r+s)=2<1. In other words,( 3 p

829+log 10

829)=2is strictly between5:5and6:5. The nearest

11. (d) The graph of4x 2

9y 2

=36is an hyperbola with vertices at( 3;0)and(3;0), and the graph ofx 2

2x+y 2

=15

is a circle of radius4centered at(1;0)and therefore crossing thex-axis at( 3;0)and(5;0). It follows that these

graphs intersect at three points.

Alternatively, if(a;b) is an intersection point, then a 2

2a+b 2

= 15 so that9a 2

18a+9b 2

= 135.

Since also4a 2

9b 2

= 36, we deduce (by adding these equations) that13a 2

18a 171 = 0. In other words, (a+3)(13a 57) = 0. Hence,a = 3 ora = 57=13. Using the equation 4a

2 9b

2

= 36, we see that in the

first caseb =0and in the second caseb = 16 p

3=13. Therefore, there are three intersection points,( 3;0)and (57=13;16

p 3=13).

12. (d) Usinga 1

=1, a direct computation givesa 2

=1,a 3

=0,a 4

=0,a 5

=2, anda 6

=1. Here, the process begins

to repeat as6is even anda 6

=1(just as2is even anda 2

=1). One deduces that for every positive integerk,a 4k

=0, a

4k+1 =2,a

4k+2

=1, anda 4k+3

=0. Thus,a j

=2precisely whenj =5;9;13;17;:::. For1j100, there are 24suchj(the last one being424+1=97).

13. (c) The number of real roots ofx 4

+jxj=10is the same as the number of intersection points of the graphs ofy=jxj

andy=10 x 4

. One easily deduces that these graphs intersect in exactly two points. Thus, there are a total of two real roots.

Alternately, ifx0andx 4

+jxj=10, thenx 4

+x=10so thatx 4

+x 10=0. This has one positive real

solution by Descartes’ Rule of Signs (and clearly0is not a solution). Ifx<0andx 4

+jxj=10, thenx 4

x=10so

thatx 4

x 10=0. Since( x) 4

( x) 10=x 4

+x 10, the equationx 4

x 10=0has exactly one negative

real root by Descartes’ Rule of Signs (alternatively, one can use that the negative real roots ofx 4

x 10=0are

precisely minus the positive real roots ofx 4

+x 10=0). Thus, again, there are a total of two real roots.

14. (e) Factoring the left-hand side of the equationx 2

xy+x y =0, we obtain(x+1)(x y) =0. The latter

is satisfied by all points(x;y)satisfying at least one ofx+1 = 0andx y = 0. In other words, the graph of x

2

xy+x y=0is composed of the two lines determined by the equationsx+1=0andx y=0(intersecting

at( 1; 1)).

15. (b) The two shaded triangles are similar, and the ratio of the length of each side of the larger triangle to the length of the corresponding side of the smaller triangle is3. It follows that the height of the larger triangle to baseADdivided

by the height of the smaller triangle to baseBCis3. Since the sum of the two heights is5, one deduces that the height

of the larger triangle is15=4and the height of the smaller triangle is5=4. It follows that the sum of the two areas is

1 2

3 15

4 +

1 2

1 5 4

= 50

8 =

25 4

=6:25:

16. (c) Let

f(x)=(1+x) 20

+x(1+x) 19

+x 2

(1+x) 18

++x 18

(1+x) 2

:

The polynomialf(x)does not change if we multiply it by1=(1+x) x. Thus,

f(x)= (1+x) x

(1+x) 20

+x(1+x) 19

+x 2

(1+x) 18

++x 18

(1+x) 2

=(1+x) 21

x 19

(1+x) 2

:

The coefficient ofx 18

when the expressionx 19

(1+x) 2

is expanded is clearly0(there is nox 18

term), so the coefficient ofx

18

inf(x)is simply the coefficient ofx 18

in(1+x) 21

. By the binomial theorem, this is

21 18

=

212019 6

=71019=1330:

Alternatively, one can use a more direct approach. The coefficient ofx 18

when the expressionx 20 k

(1+x) k

is expanded is

k k 2

=

k(k 1) 2

where2 k 20. The answer can be obtained by summing the numbers

k(k 1)=2fromk = 2tok = 20. This sum can be done more efficiently by using known formulas for P

n k

and n k=1

k 2

. (By the way, if you do not know these formulas, you might consider takingf(x)=(1+x) n

+x(1+ x)

n 1 +x

2 (1+x)

n 2

++x n 1

(1+x)+x n

and using the first argument above to derive them by looking at the coefficients ofx

n 1

andx n 2

.)

17. (e) Say the cowboys names are Bob, Dave, Kevin, and Michael. Suppose exactly two of them were shot. There are

4 2

=6ways of choosing two of them, each equally likely. Say the two shot were Dave and Kevin. The probability

that any one cowboy shot some other given cowboy is1=3. Since Dave and Kevin were the only two shot, Kevin must

have shot Dave and Dave must have shot Kevin and each of these occurred with probability1=3. Michael could have

shot Dave or Kevin, and he shot one of them with probability2=3. Similarly, Bob shot Dave or Kevin with probability 2=3. Therefore, the probability that Dave and Kevin both were shot and no one else was is

1 3

1 3

2 3

2 3

= 4 81

:

Taking into account that there are6 ways of choosing two of the four cowboys, we deduce that the probability of

exactly two cowboys being shot is64=81=8=27.

18. (d) Observe that

2N =3N N =(k

1 10

6 +k

2 10

5

++k 7

) (k 1

+k 2

10++k 7

10 6

) =(10

6 1)(k

1 k

7 )+(10

5

10)(k 2

k 6

)+(10 4

10 2

)(k 3

k 5

):

Since10 k

10 `

is divisible by9for any non-negative integerskand`, we see that the last expression above is divisible

by9. It follows that2Nand, hence,Nmust be divisible by9. One checks that the only choice of answers divisible by 9is71053290. Note that this does not prove that there arek

jas in the problem for

N =71053290(such a proof is not

required). ForN =71053290, one can takek 1

=210,k 2

=28,k 3

=28,k 4

=70,k 5

=98,k 6

=0, andk 7

=70.

19. (c) The inequalities

p n+1

p n<

1 p

4n+1 <

p n

p n 1

imply

24 X n=1

1 p

4n+1 <

24 X n=1

p n

p n 1

= p

1 p

0

+ p

2 p

1

+ p

3 p

2

+ p

4 p

3

++ p

24 p

23

= p

24<5

and

24 X

n=1 1 p

4n+1 >

24 X

n=1 p

n+1 p

n

= p

2 p

1

+ p

3 p

2

+ p

4 p

3

+ p

5 p

4

++ p

25 p

24

= p

25 p

1=5 1=4:

Hence, the answer is4.

20. (b) SettingP=(x;y), we see that

d=(x+5) 2

+(y+1) 2

+(x+1) 2

+y 2

+(x 1) 2

+(y 2) 2

+(x 1) 2

+(y 3) 2 =4x

2

+8x+4y 2

8y+42

Since(x+1) 2

0for allxwith equality precisely whenx = 1and since(y 1) 2

0for ally with equality

precisely wheny=1, the minimum value ofdis34(obtained whenx= 1andy=1).

21. (b) By the definition of the logarithm, the solutions oflog x

(5x 2) = 3forx> 2=5correspond toxfor which 5x 2=x

3

. Thus, we consider roots of the polynomialf(x)=x 3

5x+2. Sincef(x)is a cubic,f(x)has at most 3roots. One easily checks thatf( 100)<0,f(2=5)>0,f(1)<0, andf(100)>0. This implies that the graph of y =f(x)crosses thex-axis somewhere in each of the intervals( 100;2=5),(2=5;1), and(1;100). Hence, each of

these three intervals contains a root off(x). It follows thatf(x)has exactly two roots>2=5.

22. (d) The conditions in the problem imply

x 3

64x 14=(x a)(x b)(x )=x 3

(a+b+)x 2

+ ab:

Hence,a+b+=0. Also, sincea,b, andare roots ofx 3

64x 14, we obtain a

3

=64a+14; b 3

=64b+14; and 3

=64+14:

Thus,

a 3

+b 3

+ 3

=64(a+b+)+314=640+314=42:

23. (d) Multiplying the original equation by2xy, we wish to determine the number ofnon-zerointeger pairsx;y such

that2x+2y=xy. This equation is equivalent to(x 2)(y 2)=4. Thus,x 2must be plus or minus a divisor

of4 different from 2. Such a choice, sayd, for x 2uniquely determinesx(namely,x =d+2) and uniquely

determinesy(sincey 2=4=d, we havey=(4=d)+2). The number of choices forx 2is5(they are1,2, and 4). Therefore, the answer is5.

24. (a) It suffices to compute the ratio of the area ofAEDto the area ofECD. Lethdenote their common height

from vertexD. The conditions in the problem imply\ACD=30 Æ

andACDis isosceles. Hence,\CAD=30 Æ

and\ADC=120 Æ

. By the Law of Sines,

AC p

3=2 =

AC sin120

Æ =

CD sin30

Æ =

CD 1=2

so thatAC=CD p

3. Thus,

area ofAED

area ofECD =

1 2

h(AC CD) 1

2 hCD

= CD

p 3 CD CD

= p

3 1:

25. (a) Observe that when the product

1+

1 2

+ 1 2

2 +

1 2

3 +

1+

1 3

+ 1 3

2 +

1 3

3 +

is expanded, one obtains the sum of the reciprocals of all numbers of the form2 k

3 `

wherekand`are non-negative

integers. This then is precisely the sum asked for in the problem. Each of the factors in the product above is a geometric series, the first equals1=(1 (1=2)) = 2and the second equals1=(1 (1=3)) = 3=2. Therefore, the

answer is23=2=3. (Note that the product of two converging series cannot always be multiplied together

term-by-term and rearranged as above; however, if the series are converging geometric series this can be done. In fact, as long as the series are converging series with positive terms, such term-by-term multiplication and rearrangement does not alter the value of the product.)

26. (b) Observe that

1+

p 5

3

=2+ p

5 and

1 p

5

3 =2

p 5:

Hence,

3 q

2+ p

5+ 3 q

2 p

5= 1+

p 5 2

+ 1

p 5 2

=1:

A more direct approach can be given as follows. Set = 3 p

2+ p

5+ 3 p

2 p

5. Clearly,is real. Noting that (2+

p 5)(2

p

5)= 1, we obtain

3

=

3 q

2+ p

5

3 +3

3 q

2+ p

5

2

3 q

2 p

5

+3

3 q

2+ p

5

3 q

2 p

5

2 +

3 q

2 p

5

3

=(2+ p

5)+3

3 q

2+ p

5

( 1)+3

3 q

2 p

5

( 1)+(2 p

5)=4 3 :

Thus,is a root ofx 3

+3x 4. Observe that1is clearly a root. In particular,x 1is a factor ofx 3

+3x 4. Since x

3

+3x 4=(x 1)(x 2

+x+4)and the roots ofx 2

+x+4are imaginary (by the quadratic formula), we deduce

that=1.

27. (c) We consider the5 points along the diameter and the7 points along the half-circular arc separately. Two line

segments can intersect at an interior point of the semi-circle in each of the following three ways: (i) the line segments have all4endpoints on the arc, (ii) each line segment has one endpoint on the diameter and one endpoint on the arc,

and (iii) one line segment has both endpoints on the arc and the other has1endpoint on the diameter and1on the arc.

In each case, a unique intersection point is determined by the4endpoints of the two line segments (in other words,

if one knows these4endpoints, then the two line segments can be drawn in exactly one way so as to intersect at an

interior point of the semi-circle). In case (i), the4endpoints can be selected in 7 4

=35ways. Thus, (i) gives rise to 35interior points. In case (ii), the4endpoints can be selected in

7 2

5 2

=2110=210ways, producing210

additional interior points. In case (iii), the4endpoints can be selected in 7 3

5 1

=355=175ways, producing 175more interior points. Hence, there are a total of35+210+175=420interior points where line segments intersect.

28. (b) For1 j 4, letS

j denote the set of positive integers consisting of

j digits in base4. For example,S 1

= f1;2;3g. The numbers inS

1 [S

2 [S

3 [S

4are precisely the numbers from

1to255=4 4

1. Observe that asm

varies over the elements inS

j, the value of

s(m)varies over products of the formd 1

d 2

d

j where each d

jis a base 4digit (i.e., some number from the setf0;1;2;3g). We deduce that

X

m2Sj

s(m)=(0+1+2+3) j

=6 j

:

Thus,

255 X

m=1

s(m)=6+6 2

+6 3

+6 4

=6(1+6+36+216)=6259=1554:

29. (d) Drop a perpendicular fromP to sideAB, and call the intersection pointR. SetAR=xandBR=y. Similarly,

drop a perpendicular fromPto sideAD, and call the intersection pointS. SetAS=u, andDS=v. The conditions

in the problem imply

y 2

+u 2

=PB 2

=4 2

=16; u 2

+x 2

=PA 2

=8 2

=64; and x 2

+v 2

=PD 2

=7 2

=49:

Hence,

PC 2

=y 2

+v 2

=(y 2

+u 2

)+(x 2

+v 2

) (u 2

+x 2

)=16+49 64=1:

E X

Y

Z C

D

B A

F 30. (a) LetX,Y, andZ be the vertices of the shaded triangle as shown.

Observe thatCEX,AFY, andBDZall have the same area, say . Similarly, quadrilateralEAYX, quadrilateralFBZY, and

quadri-lateralDCXZ all have the same area, say. Let denote the area of XYZ. SinceCFBandCAF have the same altitude drawn from Cand sinceFB=2AF, we deduce that the area ofCFBis twice the

area ofCAF. Hence,+2+=2(2+)which implies=3.

Now, we see that the areas ofAFY andCFAare exactly one third

of the areas ofXYZandABC, respectively. Therefore, the ratio of

the area ofXYZto the area ofABCis equal to the ratio of the area

ofAFY to the area ofCFA. We compute the latter. Observe first

thatAFY andCFAare similar (since\AFY =\CFAand\FAY =\FCA). By the Law of Cosines,

CF 2

=AC 2

+AF 2

2ACAFos\CAF =(3AF)

2 +AF

2

2(3AF)AFos60 Æ =7AF

2 :

Since the ratio of the squares of the lengths of two corresponding sides of similar triangles is the ratio of the areas of the triangles, we deduce that the ratio of the area ofAFY to the area ofCFAis1=7.

S

OLUTIONS

TO

USC’

S

2000 H

IGH

S

CHOOL

M

ATH

C

ONTEST

1. (a) Since2 3

3 2

x 2

=72x 2

=9800 =2 3

5 2

7 2

, we obtainx 2

=5 2

7 2

=3 2

. Sincex>0, we deduce that x=57=3=35=3.

2. (b) Sincelog 2

(log 2

(log 2

16))=log 2

(log 2

4)=log 2

2=1, the answer is1.

3. (e) Using that2 600

= 64 100

,3 500

=243 100

,4 400

=256 100

,5 300

=125 100

, and6 200

= 36 100

, we see that the answer is6

200

.

4. (b) Letx=35811231and observe that0<x<10 8

. Hence,

10

14 +x 10

14

2 =

10 28

+(2x)10 14

+x 2 10

28

=

10000007162246210 14

+x 2 10

28

:

Since0<x 2

<10 16

, we see that the expressionx 2

does not effect the leading 12 digits of the integer in the numerator. Thus,1:00000035811231

2

=1:000000716224:::. Hence,x=7,y=1, andz=6so thatx+y+z=14.

5. (e) Given

(x 3

x 2

5x 2)(x 4

+x 3

+kx 2

5x+2)=x 7

4x 5

14x 4

5x 3

+19x 2

4;

the coefficient ofx 2

on the left is 2k+25 2= 2k+23and the coefficient ofx 2

on the right is19. These must

be equal so that 2k+23=19. Hence,k=2.

6. (d) The 100 horizontal lines divide the plane into 101 regions. If each vertical line is considered in turn, each divides the plane into 101 additional regions. The total number of regions obtained is101

2

=10201.

7. (c) The answer is2 15

=32768since it is 2 3

5

=8 5

as well as 2 5

3

=32 3

. To see thatN cannot be smaller,

observe that the conditions in the problem implyN = a 5

for some positive integeraand that for1 a 7 the

numbera 5

is not a cube of a number different froma.

8. (d) Recall that2sinos=sin (2). Hence,

(sin15 Æ

) 2

(os15 Æ

) 2

=(sin15 Æ

os15 Æ

) 2

=

sin30 Æ 2

2

=

1 4

2

= 1 16

:

9. (a) Letx=BC. SinceDBCis a45-90-45degree triangle,DB=x. SinceABCis a30-90-60degree triangle

andtan30 Æ

=1= p

3, we deduce that 1 p 3

=tan30 Æ

= BC AB

= BC AD+DB

= x 2+x

:

Therefore,2+x=x p

3, from which we deduce thatBC=x=2=( p

3 1)= p

3+1.

10. (b) Since9 3

=729and10 3

=1000, we see that 3 p

829=9+rwhere0<r<1. Since10 2

=100and10 3

=1000,

we see thatlog 10

829=2+swhere0<s<1. Thus, 3

p

829+log 10

829 2

=

(9+r)+(2+s) 2

=

11+(r+s) 2

=5:5+t;

where0 <t =(r+s)=2<1. In other words,( 3 p

829+log 10

829)=2is strictly between5:5and6:5. The nearest

11. (d) The graph of4x 2

9y 2

=36is an hyperbola with vertices at( 3;0)and(3;0), and the graph ofx 2

2x+y 2

=15

is a circle of radius4centered at(1;0)and therefore crossing thex-axis at( 3;0)and(5;0). It follows that these

graphs intersect at three points.

Alternatively, if(a;b) is an intersection point, then a 2

2a+b 2

= 15 so that9a 2

18a+9b 2

= 135.

Since also4a 2

9b 2

= 36, we deduce (by adding these equations) that13a 2

18a 171 = 0. In other words, (a+3)(13a 57) = 0. Hence,a = 3 ora = 57=13. Using the equation 4a

2 9b

2

= 36, we see that in the

first caseb =0and in the second caseb = 16 p

3=13. Therefore, there are three intersection points,( 3;0)and (57=13;16

p 3=13).

12. (d) Usinga 1

=1, a direct computation givesa 2

=1,a 3

=0,a 4

=0,a 5

=2, anda 6

=1. Here, the process begins

to repeat as6is even anda 6

=1(just as2is even anda 2

=1). One deduces that for every positive integerk,a 4k

=0, a

4k+1 =2,a

4k+2

=1, anda 4k+3

=0. Thus,a j

=2precisely whenj =5;9;13;17;:::. For1j100, there are 24suchj(the last one being424+1=97).

13. (c) The number of real roots ofx 4

+jxj=10is the same as the number of intersection points of the graphs ofy=jxj

andy=10 x 4

. One easily deduces that these graphs intersect in exactly two points. Thus, there are a total of two real roots.

Alternately, ifx0andx 4

+jxj=10, thenx 4

+x=10so thatx 4

+x 10=0. This has one positive real

solution by Descartes’ Rule of Signs (and clearly0is not a solution). Ifx<0andx 4

+jxj=10, thenx 4

x=10so

thatx 4

x 10=0. Since( x) 4

( x) 10=x 4

+x 10, the equationx 4

x 10=0has exactly one negative

real root by Descartes’ Rule of Signs (alternatively, one can use that the negative real roots ofx 4

x 10=0are

precisely minus the positive real roots ofx 4

+x 10=0). Thus, again, there are a total of two real roots.

14. (e) Factoring the left-hand side of the equationx 2

xy+x y =0, we obtain(x+1)(x y) =0. The latter

is satisfied by all points(x;y)satisfying at least one ofx+1 = 0andx y = 0. In other words, the graph of x

2

xy+x y=0is composed of the two lines determined by the equationsx+1=0andx y=0(intersecting

at( 1; 1)).

15. (b) The two shaded triangles are similar, and the ratio of the length of each side of the larger triangle to the length of the corresponding side of the smaller triangle is3. It follows that the height of the larger triangle to baseADdivided

by the height of the smaller triangle to baseBCis3. Since the sum of the two heights is5, one deduces that the height

of the larger triangle is15=4and the height of the smaller triangle is5=4. It follows that the sum of the two areas is

1 2

3 15

4 +

1 2

1 5 4

= 50

8 =

25 4

=6:25:

16. (c) Let

f(x)=(1+x) 20

+x(1+x) 19

+x 2

(1+x) 18

++x 18

(1+x) 2

:

The polynomialf(x)does not change if we multiply it by1=(1+x) x. Thus,

f(x)= (1+x) x

(1+x) 20

+x(1+x) 19

+x 2

(1+x) 18

++x 18

(1+x) 2

=(1+x) 21

x 19

(1+x) 2

:

The coefficient ofx 18

when the expressionx 19

(1+x) 2

is expanded is clearly0(there is nox 18

term), so the coefficient ofx

18

inf(x)is simply the coefficient ofx 18

in(1+x) 21

. By the binomial theorem, this is

21 18

=

212019 6

=71019=1330:

Alternatively, one can use a more direct approach. The coefficient ofx 18

when the expressionx 20 k

(1+x) k

is expanded is

k k 2

=

k(k 1) 2

where2 k 20. The answer can be obtained by summing the numbers

k(k 1)=2fromk = 2tok = 20. This sum can be done more efficiently by using known formulas for P

n k

and n k=1

k 2

. (By the way, if you do not know these formulas, you might consider takingf(x)=(1+x) n

+x(1+ x)

n 1 +x

2 (1+x)

n 2

++x n 1

(1+x)+x n

and using the first argument above to derive them by looking at the coefficients ofx

n 1

andx n 2

.)

17. (e) Say the cowboys names are Bob, Dave, Kevin, and Michael. Suppose exactly two of them were shot. There are

4 2

=6ways of choosing two of them, each equally likely. Say the two shot were Dave and Kevin. The probability

that any one cowboy shot some other given cowboy is1=3. Since Dave and Kevin were the only two shot, Kevin must

have shot Dave and Dave must have shot Kevin and each of these occurred with probability1=3. Michael could have

shot Dave or Kevin, and he shot one of them with probability2=3. Similarly, Bob shot Dave or Kevin with probability 2=3. Therefore, the probability that Dave and Kevin both were shot and no one else was is

1 3

1 3

2 3

2 3

= 4 81

:

Taking into account that there are6 ways of choosing two of the four cowboys, we deduce that the probability of

exactly two cowboys being shot is64=81=8=27.

18. (d) Observe that

2N =3N N =(k

1 10

6 +k

2 10

5

++k 7

) (k 1

+k 2

10++k 7

10 6

) =(10

6 1)(k

1 k

7 )+(10

5

10)(k 2

k 6

)+(10 4

10 2

)(k 3

k 5

):

Since10 k

10 `

is divisible by9for any non-negative integerskand`, we see that the last expression above is divisible

by9. It follows that2Nand, hence,Nmust be divisible by9. One checks that the only choice of answers divisible by 9is71053290. Note that this does not prove that there arek

jas in the problem for

N =71053290(such a proof is not

required). ForN =71053290, one can takek 1

=210,k 2

=28,k 3

=28,k 4

=70,k 5

=98,k 6

=0, andk 7

=70.

19. (c) The inequalities

p n+1

p n<

1 p

4n+1 <

p n

p n 1

imply

24 X n=1

1 p

4n+1 <

24 X n=1

p n

p n 1

= p

1 p

0

+ p

2 p

1

+ p

3 p

2

+ p

4 p

3

++ p

24 p

23

= p

24<5

and

24 X

n=1 1 p

4n+1 >

24 X

n=1 p

n+1 p

n

= p

2 p

1

+ p

3 p

2

+ p

4 p

3

+ p

5 p

4

++ p

25 p

24

= p

25 p

1=5 1=4:

Hence, the answer is4.

20. (b) SettingP=(x;y), we see that

d=(x+5) 2

+(y+1) 2

+(x+1) 2

+y 2

+(x 1) 2

+(y 2) 2

+(x 1) 2

+(y 3) 2 =4x

2

+8x+4y 2

8y+42

Since(x+1) 2

0for allxwith equality precisely whenx = 1and since(y 1) 2

0for ally with equality

precisely wheny=1, the minimum value ofdis34(obtained whenx= 1andy=1).

21. (b) By the definition of the logarithm, the solutions oflog x

(5x 2) = 3forx> 2=5correspond toxfor which 5x 2=x

3

. Thus, we consider roots of the polynomialf(x)=x 3

5x+2. Sincef(x)is a cubic,f(x)has at most 3roots. One easily checks thatf( 100)<0,f(2=5)>0,f(1)<0, andf(100)>0. This implies that the graph of y =f(x)crosses thex-axis somewhere in each of the intervals( 100;2=5),(2=5;1), and(1;100). Hence, each of

these three intervals contains a root off(x). It follows thatf(x)has exactly two roots>2=5.

22. (d) The conditions in the problem imply

x 3

64x 14=(x a)(x b)(x )=x 3

(a+b+)x 2

+ ab:

Hence,a+b+=0. Also, sincea,b, andare roots ofx 3

64x 14, we obtain a

3

=64a+14; b 3

=64b+14; and 3

=64+14:

Thus,

a 3

+b 3

+ 3

=64(a+b+)+314=640+314=42:

23. (d) Multiplying the original equation by2xy, we wish to determine the number ofnon-zerointeger pairsx;y such

that2x+2y=xy. This equation is equivalent to(x 2)(y 2)=4. Thus,x 2must be plus or minus a divisor

of4 different from 2. Such a choice, sayd, for x 2uniquely determinesx(namely,x =d+2) and uniquely

determinesy(sincey 2=4=d, we havey=(4=d)+2). The number of choices forx 2is5(they are1,2, and 4). Therefore, the answer is5.

24. (a) It suffices to compute the ratio of the area ofAEDto the area ofECD. Lethdenote their common height

from vertexD. The conditions in the problem imply\ACD=30 Æ

andACDis isosceles. Hence,\CAD=30 Æ

and\ADC=120 Æ

. By the Law of Sines,

AC p

3=2 =

AC sin120

Æ =

CD sin30

Æ =

CD 1=2

so thatAC=CD p

3. Thus,

area ofAED

area ofECD =

1 2

h(AC CD) 1

2 hCD

= CD

p 3 CD CD

= p

3 1:

25. (a) Observe that when the product

1+

1 2

+ 1 2

2 +

1 2

3 +

1+

1 3

+ 1 3

2 +

1 3

3 +

is expanded, one obtains the sum of the reciprocals of all numbers of the form2 k

3 `

wherekand`are non-negative

integers. This then is precisely the sum asked for in the problem. Each of the factors in the product above is a geometric series, the first equals1=(1 (1=2)) = 2and the second equals1=(1 (1=3)) = 3=2. Therefore, the

answer is23=2=3. (Note that the product of two converging series cannot always be multiplied together

term-by-term and rearranged as above; however, if the series are converging geometric series this can be done. In fact, as long as the series are converging series with positive terms, such term-by-term multiplication and rearrangement does not alter the value of the product.)

26. (b) Observe that

1+

p 5

3

=2+ p

5 and

1 p

5

3 =2

p 5:

Hence,

3 q

2+ p

5+ 3 q

2 p

5= 1+

p 5 2

+ 1

p 5 2

=1:

A more direct approach can be given as follows. Set = 3 p

2+ p

5+ 3 p

2 p

5. Clearly,is real. Noting that (2+

p 5)(2

p

5)= 1, we obtain

3

=

3 q

2+ p

5

3 +3

3 q

2+ p

5

2

3 q

2 p

5

+3

3 q

2+ p

5

3 q

2 p

5

2 +

3 q

2 p

5

3

=(2+ p

5)+3

3 q

2+ p

5

( 1)+3

3 q

2 p

5

( 1)+(2 p

5)=4 3 :

Thus,is a root ofx 3

+3x 4. Observe that1is clearly a root. In particular,x 1is a factor ofx 3

+3x 4. Since x

3

+3x 4=(x 1)(x 2

+x+4)and the roots ofx 2

+x+4are imaginary (by the quadratic formula), we deduce

that=1.

27. (c) We consider the5 points along the diameter and the7 points along the half-circular arc separately. Two line

segments can intersect at an interior point of the semi-circle in each of the following three ways: (i) the line segments have all4endpoints on the arc, (ii) each line segment has one endpoint on the diameter and one endpoint on the arc,

and (iii) one line segment has both endpoints on the arc and the other has1endpoint on the diameter and1on the arc.

In each case, a unique intersection point is determined by the4endpoints of the two line segments (in other words,

if one knows these4endpoints, then the two line segments can be drawn in exactly one way so as to intersect at an

interior point of the semi-circle). In case (i), the4endpoints can be selected in 7 4

=35ways. Thus, (i) gives rise to 35interior points. In case (ii), the4endpoints can be selected in

7 2

5 2

=2110=210ways, producing210

additional interior points. In case (iii), the4endpoints can be selected in 7 3

5 1

=355=175ways, producing 175more interior points. Hence, there are a total of35+210+175=420interior points where line segments intersect.

28. (b) For1 j 4, letS

j denote the set of positive integers consisting of

j digits in base4. For example,S 1

= f1;2;3g. The numbers inS

1 [S

2 [S

3 [S

4are precisely the numbers from

1to255=4 4

1. Observe that asm

varies over the elements inS

j, the value of

s(m)varies over products of the formd 1

d 2

d

j where each d

jis a base 4digit (i.e., some number from the setf0;1;2;3g). We deduce that

X

m2Sj

s(m)=(0+1+2+3) j

=6 j

:

Thus,

255 X

m=1

s(m)=6+6 2

+6 3

+6 4

=6(1+6+36+216)=6259=1554:

29. (d) Drop a perpendicular fromP to sideAB, and call the intersection pointR. SetAR=xandBR=y. Similarly,

drop a perpendicular fromPto sideAD, and call the intersection pointS. SetAS=u, andDS=v. The conditions

in the problem imply

y 2

+u 2

=PB 2

=4 2

=16; u 2

+x 2

=PA 2

=8 2

=64; and x 2

+v 2

=PD 2

=7 2

=49:

Hence,

PC 2

=y 2

+v 2

=(y 2

+u 2

)+(x 2

+v 2

) (u 2

+x 2

)=16+49 64=1:

E X

Y

Z C

D

B A

F 30. (a) LetX,Y, andZ be the vertices of the shaded triangle as shown.

Observe thatCEX,AFY, andBDZall have the same area, say . Similarly, quadrilateralEAYX, quadrilateralFBZY, and

quadri-lateralDCXZ all have the same area, say. Let denote the area of XYZ. SinceCFBandCAF have the same altitude drawn from Cand sinceFB=2AF, we deduce that the area ofCFBis twice the

area ofCAF. Hence,+2+=2(2+)which implies=3.

Now, we see that the areas ofAFY andCFAare exactly one third

of the areas ofXYZandABC, respectively. Therefore, the ratio of

the area ofXYZto the area ofABCis equal to the ratio of the area

ofAFY to the area ofCFA. We compute the latter. Observe first

thatAFY andCFAare similar (since\AFY =\CFAand\FAY =\FCA). By the Law of Cosines,

CF 2

=AC 2

+AF 2

2ACAFos\CAF =(3AF)

2 +AF

2

2(3AF)AFos60 Æ =7AF

2 :

Since the ratio of the squares of the lengths of two corresponding sides of similar triangles is the ratio of the areas of the triangles, we deduce that the ratio of the area ofAFY to the area ofCFAis1=7.